Question

Find the equation to the straight line which bisects, and is perpendicular to the straight line joining the points $(a,b)$ and $(a^{\prime },b^{\prime })$.

Answer 1

Slope between two given points is $m=\frac{b^{{}^{\prime }}-b}{a^{{}^{\prime }}-a}$

Then perpendicular slope is $m_1=-\frac{a^{{}^{\prime }}-a}{b^{{}^{\prime }}-b}$

The point where the line bisects is $(\frac{a+a^{{}^{\prime }}}{2},\frac{b+b^{{}^{\prime }}}{2})$.......$\left(0\right)$

Then the equation of the line with this slope is $y=\frac{a-a^{{}^{\prime }}}{b^{{}^{\prime }}-b}x+c$.......$\left(1\right)$

Given that the above straight line passes through the the point $\left(0\right)$

Then, $\frac{b+b^{{}^{\prime }}}{2}=\frac{a-a^{{}^{\prime }}}{b^{{}^{\prime }}-b}\times \left(\frac{a+a^{{}^{\prime }}}{2}\right)+c$

$\Rightarrow c=\frac{(b^{{}^{\prime }}{)}^2-b^2+(a^{\prime }{)}^2-a^2}{2(b-b^{{}^{\prime }})}$

Therefore, required line is $2x(a^{{}^{\prime }}-a)+2y(b^{{}^{\prime }}-b)=a^2-(a^{{}^{\prime }}{)}^2+b^2-(b^{{}^{\prime }}{)}^2$ (Substitute ${}^{\prime }{c}^{\prime }$ in $\left(1\right)$ and simplify)